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Musya8 [376]
3 years ago
12

A protocol is a set of rules or procedures, usually written, that should be followed in specific situations. Which of the follow

ing are examples of protocols? (Select all that apply.)
removing your hat during the playing of the national anthem
filing an injury claim with OSHA
driving on the right hand side of the road
hosting a barbeque on the Fourth of July
Engineering
1 answer:
dangina [55]3 years ago
5 0
The answer is 2 and 3
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Let suppose, you are going to develop a web-application for school management system. Then what architectural pattern will you u
Gnesinka [82]

Answer:

The architectural pattern i will use for the school management is the client-server pattern.

This pattern would consist of  a server and many clients. wherein the server component would provide services to that of the clients and its components as specified and also there would be a client request service from that of the server.

Explanation:

Solution

A school management system would always involve the client server pattern as this pattern would have a server and many clients wherein the server component would give services to that of the clients and its components as specified and also there would be a client request service from that of the server. This server would share the appropriate services to such clients and also listen to the client's requests.

Such kind of pattern would mostly be used for for the online platforms or application like that of document.

5 0
3 years ago
Explain why surface temperature increases when two bodies are rubbed against each other. What is the significance of temperature
Ronch [10]

Answer:

The surface temperature increases when two bodies are rubbed against each other due to friction.

Explanation:

No object has a perfectly even surface. So, when two bodies with uneven surfaces are rubbed against each other, they experience friction.

Friction is a resistance experienced by the two bodies when they are moved against each other.

The friction between the two surfaces, converts the kinetic energy of the movement to the thermal energy.

Thus, resulting in rise in the surface temperature of the two bodies.

Therefore, when two bodies are rubbed against each other, the surface temperature increases due to friction.

7 0
3 years ago
An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0
3 years ago
How do you check battery state of charge with a voltmeter
cupoosta [38]

Answer:

Depends on the battery and the current type.

Is it AC or DC?

Explanation:

Could you mark as brainiest.

I need it for my account

Thank you! :)

8 0
2 years ago
A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f
dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14,  pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, head\ loss  = \frac{f L V^2}{( 2 g D)}

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

V = \frac{Q}{A}

    = \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s

obtained Darcy friction factor  

calculate Reynold number (Re) ,

Re = \frac{\rho V D}{\mu}

where,\rho = density of water

\mu = Dynamic viscosity of water at 15 degree  C = 0.001 Ns/m2

so reynold number is

Re = \frac{1000\times 2.015\times 0.318}{0.001}

            = 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is  0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}

HL = 1.64 m

7 0
3 years ago
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